How do you factor #3a^2 + 14a − 24#?

2 Answers
Apr 6, 2016

Answer:

#(3a - 4)(a + 6)#

Explanation:

We know there must be a #3a# and #a# in the brackets because they are the only factors of #3#.

You then use trial and error to find a the other numbers which multiply together to make #24# and add together (multiplied by their opposite coefficients of #a#) to make #14#, which you find are #6# and #-4#.

Apr 6, 2016

Answer:

(3a - 4)(a + 6)

Explanation:

Use the systematic, non-guessing new AC Method (Socratic Search).
#y = 3a^2 + 14a - 24 =# 3(a + p)(a + q)
Converted trinomial #y' = a^2 + 14a - 72 =# (a + p')(a + q')
p' and q' have opposite signs because ac < 0.
Factor pairs of (ac = -72) --> (-3,24) (-4, 18). This sum is 14 = b. Then,
p' = -4 and q' = 18.
Back to trinomial y, # p = (p')/a = - 4/3# and #q = (q')/a = 18/3 = 6#
Factored form: #y = 3(a - 4/3)(a + 6) = (3a - 4)(a + 6)#